hw5solutions - IEOR 130 Spring 2007 Prof Leachman Solutions...

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IEOR 130 Spring, 2007, Prof. Leachman Solutions to HW 5 1. c 1 = 4.5, c 2 = 0.5 t p t 1-F(t) denominator numerator G(t) Non-availability= G(t)/168 1 .15 .85 1.0 1.10 1.10 0.00655 2 .20 .65 1.85 1.90 1.027 0.00611 3 .25 .40 2.50 2.90 1.16 0.00690 4 .25 .15 2.90 3.90 1.34 0.00798 5 .15 0 3.05 4.50 1.48 0.00881 Replace O ring every 2 weeks. Note G(t) above is expressed as hours of down time per weeks of cycle length, so to express true non-availability we would have to convert units to hours of down time divided by hours of cycle length, i.e., non-availability equals G(t)/168. 2. (a) c 1 = 200, c 2 = 50 t p t sum k*p k 1-F(t) denominator numerator G(t) 1 .08 .08 . 92 1.0 62 62.0 2 .12 .32 .80 1.92 80 41.7 3 .16 .80 .64 2.72 104 38.2 4 .24 1.76 .40 3.36 140 41.7 t=3 gives the lowest cost rate. (b) The production rate, λ = 10, A=0.5, DY = (0.5)*exp{-A*(delta D)} Revenue = 2*100*(0.50)*exp{-A*(delta D)} Delta revenue = 100*[1 - exp{-A(delta D)}] per wafer If we PM every day, delta D = 0.1. Hence Delta revenue = 100*[1 - exp{-.05}] ~ 100*(.05) If we PM every 3 days, delta D = 0.3. Hence Delta revenue = 100*[1 - exp{-.15}] ~ 100*(.14) We should modify the G(t) function to be G*(t) = G(t) + { λ *{Delta revenue}*denominator} / denominator Computing for the alternative values of t, we find t G*(t) 1 62.0 + 50 = 112.0 3 38.2 + 140 = 178.2 So we should do the PM every day in this case.
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3. c 1 = 6.5, c 2 variable t p t 1-F(t) 1 .05 .95 2 .05 .90 3 .05 .85 4 .05 .80 5 .05 .75 6 .05 .70 7 .05 .65 8 .05 .60 9 .05 .55 10 .05 .50 11 .05 .45 12 .05 .40 G(t) = { c 1 - (c 1 - c 2 )(1 - F(t)) } / [ sum k=1 to t { k*p k } + t*(1-F(t)) ] G(1) = {6.5 - (6.5 - c 2 )(0.95)} / 1 = 0.95*c 2 + 0.325 G(4) = {6.5 - (6.5 - c 2 )(0.80)} / [ 4(0.80) + (0.05)(1+2+3+4) ] = 0.216*c 2 + 0.351 G(12) = {6.5 - (6.5 - c 2 )(0.40)} / [ 12(0.40) + (0.05)(1+2+3+...+12) ] = {6.5 - (6.5 - c 2 )(0.40)} / [ 4.8 + (0.05)*78 ] = (0.40*c 2 + 0.6*6.5) / 8.7 = 0.046*c 2 + 0.448 Cost rate if let fail: 6.5 / { sum k=1 to t k*p k } = 6.5 / { .05 * sum k=1 to t p k } = 6.5 / { (.05)(20)(21) / 2 } = 0.619 0.95*c 2 + 0.325 < 0.216*c 2 + 0.351 0.734*c 2 < 0.026 c 2 < 0.0354 hours or about 2 minutes 0.216*c 2 + 0.351 < 0.046*c 2 + 0.446 0.17*c 2 < 0.095 c 2 < 0.5588 hours or about 34 minutes 0.046*c 2 + 0.446 < 0.619 0.046*c 2 < 0.173 c 2 < 3.76 hours Recap: If 0 < c 2 <= 0.0354 hours, do weekly PM
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  • Spring '08
  • Leachman
  • Probability theory, Mathematics in medieval Islam, Replacements, Rate function

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